At time t in seconds, a particle's distance s(t), in centimeters, from a point is given by s(t) = 4 + 3 sin t. What is the average velocity of the particle from t = ð/3 to t = 7ð/3?
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To find the average velocity of the particle from t = π/3 to t = 7π/3, we need to calculate the total change in distance (Δs) and divide it by the total change in time (Δt).
First, let's find the distance at t = π/3 and t = 7π/3:
- Plug in t = π/3 into the equation s(t) = 4 + 3 sin t:
s(π/3) = 4 + 3 sin(π/3) = 4 + 3 * (√3/2) = 4 + 3√3/2
- Plug in t = 7π/3 into the equation s(t) = 4 + 3 sin t:
s(7π/3) = 4 + 3 sin(7π/3) = 4 + 3 * (-√3/2) = 4 - 3√3/2
The total change in distance (Δs) is the difference between these two distances:
Δs = [4 - 3√3/2] - [4 + 3√3/2]
= [4 - 3√3/2 - 4 - 3√3/2]
= -6√3/2
Next, let's find the change in time (Δt):
Δt = (7π/3) - (π/3)
= (7π - π)/3
= 6π/3
= 2π
Now we can calculate the average velocity:
Average velocity = Δs / Δt
= -6√3/2 / 2π
= -3√3/π
Therefore, the average velocity of the particle from t = π/3 to t = 7π/3 is -3√3/π centimeters per second.